Question

1. In a non-transparent bag there are 5 equally-weighted balls, each of them is marked by a number 1,2,3,4 and 5. Now we randomly pick 3 balls out of the bag and we note X as the smallest number among the 3 balls picked. The order is of no importance.(10 pt)
   1. What is the statistic distribution of X?
   2. Compute E(X) and also .

2. Suppose that a factory products the same product using 3 different workshops A, B and C, the producing capacity ratio of these workshops are A: B: C=3:2:1. The failure rates of product produced by these workshops are 8% for workshop A, 9% for workshop B and 12% for workshop C. Now we randomly select a product produced by this factory. (10 pt)
   1. Using conditional probability theory, what is the probability that this product selected is a failure one ?
   2. If the product select is indeed a failure one, what is the probability that it’s produced by workshop A?

Answer 1

1.

a.

The support of X are 1, 2, 3 (X cannot be 4 or 5 as there are three balls to picked and at least one of the number will be less than 4 and 5)

Number of ways to pick three balls from five if order is of no importance = 5C3 = 5! / ((5-3)! 3!) = (5 * 4) / (1 * 2) = 10

If suppose a ball numbered 1 is already picked, Number of ways to pick remaining 2 balls from 4 = 4C2 = 6

P(X = 1) = Number of ways to pick balls so that ball number 1 is there / Total number of ways to pick the ball

= 6/10 = 0.6

If suppose a ball numbered 2 is already picked and ball number 1 is not picked up, Number of ways to pick remaining 2 balls from 3 = 3C2 = 3

P(X = 2) = Number of ways to pick balls so that ball number 1 is not there and ball number 2 is there / Total number of ways to pick the ball

= 3/10 = 0.3

If suppose a ball numbered 3 is already picked and ball number 1, 2 are not picked up, Number of ways to pick remaining 2 balls from 2 = 2C2 = 1

P(X = 3) = Number of ways to pick balls so that ball number 1, 2 are not there and ball number 3 is there / Total number of ways to pick the ball

= 1/10 = 0.1

The statistic distribution of X is,

X	P(X)
1	0.6
2	0.3
3	0.1

b.

E(X) = 1 * 0.6 + 2 * 0.3 + 3 * 0.1 = 1.5

2.

P(A) = 3 / (3+2+1) = 1/2

P(B) = 2 / (3+2+1) = 1/3

P(C) = 1 / (3+2+1) = 1/6

P(F | A) = 0.08

P(F | B) = 0.09

P(F | C) = 0.12

a.

probability that this product selected is a failure one = P(F) =

= P(F | A) P(A) + P(F | B) P(B) + P(F | C) P(C)

= 0.08 * (1/2) + 0.09 * (1/3) + 0.12 * (1/6)

= 0.09

b.

If the product select is indeed a failure one, probability that it’s produced by workshop A

= P(A | F)

= P(F | A) P(A) / P(F) (Bayes Theorem)

= 0.08 * (1/2) / 0.09

= 0.4444444

= 4/9